Answer to Question #193763 in Mechanics | Relativity for jay

Question

In a plane electromagnetic wave, the electric field (in V/m) is given by equation

Ez= 6.0 sin [2π(2· 10¹⁰ t + 500 x)] where x is in metre and t in second.

(a) What is the direction of propagation of the wave?

(b) What is the rms value of electric field?

(c) Find the wavelength and frequency of the wave.

(d) Write the expression for the magnetic field.

Answer 1

$E_z=6.0sin[2{\pi}\ (2\times10^{10}t +500x)]$

Now transferring this equation in general form we get =$E_z=$ $6.0sin[(4{\pi}\times 10^{10}t+500x)]$

(a) The direction of propagation of wave can be determined very easily -

If we write general form of equation which is $E_z=E_osin [2{\pi}(\omega$t+k$x)]$

now $\omega$t + k$x$ in this case is equal to =$4{\pi}\times10^{10}t+500x$ =0

now, differentiating the above equation with respect to t , we get -

$=4{\pi}\times10^{10}+500\dfrac{dx}{dt}=0$

now we can see that $\dfrac{dx}{dt}$ turns out to be negative , it means that wave is moving in $(-x)$ direction .

so direction of wave is confirmed it is moving in negative $-x\$ , direction.

(b) The rms value of electric field , in this case $E_{rms}$ can be calculated very easily $E_{rms}$ =$\dfrac{E_{max}}{\sqrt2}$

$E_{max}=0.6$ , so our $E_{rms}=$ 0.526

(c) Wavelength and frequency of wave can be determined very easily -

Now we know that k = $\dfrac{2\pi}{\lambda}$, k=500 given in the equation, ${\lambda}=\dfrac{2{\pi}}{500}$ = 0.012 m

Now for $\omega$ =2${\pi}f$ =4${\pi}\times 10^{10}$

$f=2\times10^{10}Hz$

(d) The expression of magnetic field equation can be written as , the terms frequency and wavelength are same in both electric and magnetic field so not much of manipulation is there

$B_Z$ =6.0 $sin[2{\pi}(2\times10^{10}t+500x)]$